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Summary: This module explains rotational kinetic energy and inertia in a format that is accessible to blind students.
This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.
This module explains rotational kinetic energy and inertia in a format that is accessible to blind students.
In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:
The minimum prerequisites for understanding the material in these modules include:
I recommend that you open another copy of this document in a separate browser window and use the following links to easily find and view the material listed below while you are reading about it.
I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .
What do we mean by rotational kinetic energy and rotational inertia?
In an earlier module, you learned of Newton's first law, which can be paraphrased something like the following:
Newton's first law
Every body (that has mass) in a state of rest tends to remain at rest. Similarly, every body (that has mass) in a state of motion tends to remain in motion in a straight line. In both cases, the body tends to remain in its current state unless compelled to change its state by external forces acting upon it.
This law is sometimes referred to as the "Law of Inertia" -- objects that have mass don't like to change their velocity.
Newton's second law
Newton's second implies (once again paraphrasing) that the tendency to remain in the state of rest or motion depends on the amount of mass possessed by the body.
The greater the mass, the greater must be the external force required to cause the body to change its state by a given amount. You will recognize this as being characterized by the equation that tells us that acceleration (change of velocity) is proportional to the applied force and inversely proportional to the amount of mass.
Rotating rigid bodies
Similar considerations also apply to the rotation of rigid bodies. In the case of rigid-body rotation, however, it isn't simply the amount of mass that is important. The geometrical distribution of that mass also has an impact on the reluctance of a rotating body to change its state. We refer to this as rotational inertia , which is sometimes called the moment of inertia .
Rotational inertia or moment of inertia?
Rotational inertia and moment of inertia are simply two names that mean the same thing (some authors prefer one, other authors prefer the other)
I don't have a preference for either, but for consistency with the textbook currently being used for introductory physics courses at the college where I teach, I will use the term rotational inertia instead of moment of inertia .
An example
For example, the same amount of mass can be used to create
When rotated about its central axis, either disk has a greater rotational inertia than the cylinder, and the disk with most of its mass concentrated around the outer circumference has a greater rotational inertia than the other two.
The rotational inertia increases as the mass is moved further from the axis of rotation, and the effect is proportional to square of that distance. As a result, given that the total mass is the same in all three cases, a greater external force is required to cause either disk to change its rotational state than is required to cause the cylinder to change its rotational state.
Work and kinetic energy
You also learned in an earlier module that work must be done on an object to cause it to have kinetic energy, and the kinetic energy possessed by an object is proportional to one-half of the product of the object's mass and the square of its velocity.
KE = (1/2)*m*v^2
where
The rotational kinetic energy of a rotating body
When I was a youngster, I learned the hard way that a rotating rigid object has rotational kinetic energy. I had my bicycle turned upside down resting on the seat and the handle bars with the rear wheel turning very fast. I was using a long thin triangular file to chip mud off of the bicycle. I accidentally allowed one end of the file to come in contact with the tread on the spinning bicycle tire and ended up with a file sticking in the palm of my hand. Although I didn't know the technical term for rotational kinetic energy at the time, I did learn what rotational kinetic energy can do.
In principle, at least, we could calculate the rotational kinetic energy possessed by that spinning bicycle wheel by
That would be a difficult computation. We need a simpler way to express the rotational kinetic energy of a rotating rigid body.
A simpler way
There is a simpler way that is based on the tangential speed of each particle of mass and the angular velocity of the rotating object.
You should recall that when the angular velocity is expressed in radians per second, the tangential speed of a point on the circumference of a circle is given by
v = r * w
where
Thus, the tangential speed of our hypothetical particle of mass is equal to the product of the distance of that particle from the center of rotation (the axle on my upturned bicycle) and the angular velocity of the wheel.
Terminology
The convention is to use the Greek letter omega to represent angular velocity, but I decided to use the "w" character because
Back to the bicycle wheel
Therefore, if we consider the bicycle wheel to be made up of an extremely large number of particles of mass, each located at a fixed distance from the axle, the kinetic energy of each particle would be given by
KEr = (1/2)*m*v^2, or
KEr = (1/2)*m*(r*w)^2, or
KEr = (1/2)*m*(r^2)*(w^2)
where all of the terms in this equation were defined earlier except for
The total rotational kinetic energy of the bicycle wheel
Then the total rotational kinetic energy of the bicycle wheel would be
KErt = (1/2)*(sum from i=0 to i=N(mi*ri^2))*w^2
where
Summation
It is conventional to use the Greek letter sigma to represent the sum with subscripts and superscripts providing the limits of the sum. However, since your Braille display probably won't display the Greek letter sigma with subscripts and superscripts, we will have to settle for something like "sum from i=0 to i=N" to mean the same thing.
Factoring out like terms
Note that then entire wheel rotates at the same angular velocity, so we can (and did) factor the (1/2) and the w^2 out of the summation equation given above.
Integral calculus is the key
Regardless of how difficult it may seem to you to perform the summation given above, when you complete a course in integral calculus, you will have learned how to do that sort of thing for a variety of geometric shapes such as solid cylinders, hollow cylinders, solid spheres, hollow spheres, squares, rectangles, rods, etc. As a result, engineering and physics handbooks contain tables with this sort of information for a variety of common geometrical shapes.
(There is also such a table at http://en.wikipedia.org/wiki/List_of_moments_of_inertia , but if you are a blind student, your accessibility equipment may not be able to read it reliably.)
Getting back to the earlier equation for rotational kinetic energy , the quantity in parentheses cannot change for a given geometric shape. The distance between each mass particle and the axis of rotation stays the same for a rigid body, and the mass of each mass particle doesn't change. It is conventional to give the quantity in parentheses the symbol I (upper-case "I") and refer to it as either the rotational inertia , or the moment of inertia .
Therefore, using the terminology from the earlier equation for rotational kinetic energy ,
Facts worth remembering -- Rotational Inertia
I = sum from i=0 to i=N(mi*ri^2)
where
Translational versus rotational kinetic energy
Given that information, let's form an analogy between translational kinetic energy from an earlier module and rotational kinetic energy from this module.
Facts worth remembering -- Translational and Rotational Kinetic Energy
Kt = (1/2)*m*v^2
Kr = (1/2)*I*w^2
where
Comparing the terms
When we compare the terms in the two expressions, we see that angular velocity in one case is analogous to translational velocity in the other case.
We also see that the rotational inertia in one expression is analogous to the mass in the other expression.
As I explained earlier, mass is an absolute in translational terms. The translational inertia depends directly on the amount of mass.
However, mass is not an absolute in rotational terms. The rotational inertia for rotation depends not only upon the amount of mass involved, but also on how that mass is geometrically distributed within the object relative to the axis of rotation.
A measure of inertia
In translational terms, mass is a measure of the inertia of an object, or how difficult it is to cause the object to change its translational velocity.
In rotational terms, for a rigid rotating object, the rotational inertia ( I ), is a measure of how hard it is to cause the object to change its angular velocity.
Finding the rotational inertia
Here are a few tips on how you might go about finding the rotational inertia of an object.
Facts worth remembering -- Finding the rotational inertia
The rotational axis is very important
The rotational inertia for an object depends heavily on the location of the axis of rotation. For example, some vehicles have doors on the back that are hinged on one side. Other vehicles have doors on the back that are hinged at the top. Given a door that has a rectangular shape, but which is not a square, the rotational inertia when the door is hinged on the side would be different from when the door is hinged at the top.
Assuming that both doors have the same mass, and are fastened to the vehicle with the same orientation, the center of mass for one arrangement would be further from the hinge than for the other arrangement. The arrangement for which the center of mass is further from the hinge would have the greater rotational inertia.
A simple experiment
Pick up an eight-foot piece of 2x4 lumber, grasp it near one end, and try swinging it like a baseball bat. You should find that to be relatively difficult because it has a large rotational inertia when rotated around its end. (It also has a lot torque due to gravity when supported only at the end. Torque will be the topic for a future module.)
Then grasp it in the center and rotate it as far as you can without hitting your body. You should find that to be somewhat easier because it has a smaller rotational inertia when rotated around its center than when rotated around its end.
I will have more to say about this later in this module.
It is possible to determine the rotational inertia of an object about any axis if we can determine the rotational inertia of that same object about a parallel axis that goes through the center of mass of the object.
I will explain this in much more detail in the dumbbell scenario later in this module.
Facts worth remembering -- The parallel axis theorem
The total rotational inertia of an object about a chosen axis is
We can express this theorem in equation form as
Itotal = M*D^2 + Icm
where
In this section, I will attempt to describe some simple geometric shapes and provide the formula for the rotational inertia for each shape. This information is largely based on information gleaned from http://en.wikipedia.org/wiki/List_of_moments_of_inertia .
Terminology
Unless I specify otherwise, the axis of rotation will be the axis of symmetry, such as at the center of a wheel. Also, unless I specify otherwise,
Facts worth remembering -- Examples of rotational inertia
Thin hollow cylindrical shape or hoop
Think of a can of beans without the beans and without the end caps.
I = m*r^2
Solid cylinder or disk
I = (1/2)*m*r^2
Thick-walled cylindrical tube with open ends, of inner radius r1, and outer radius r2
I = (1/2)*m*(r1^2 + r2^2)
Thin rectangular plate of height h and width w with axis of rotation in the center, perpendicular to the plate
I = (1/12)*m*(h^2 + w^2)
Solid sphere
I = (2/5)*m*r^2
Then hollow spherical shell
I = (2/3)*m*r^2
Thin rod of length L
Axis of rotation is perpendicular to the end of the rod.
I = (1/3)*m*L^2
Thin rectangular plate of width L and height H
Axis of rotation is along the edge of the plate parallel to the H dimension and perpendicular to the width L.
I = (1/3)*m*L^2
Thin rod of length L
Axis of rotation is through the center of the rod.
I = (1/12)*m*L^2
Thin rectangular plate of width L and height H
Axis of rotation is along the center of the plate parallel to the H dimension perpendicular to the width L.
I = (1/12)*m*L^2
I will apply some of what we have learned to several different scenarios in this section.
Returning to the earlier example, pick up an eight-foot piece of 2x4 lumber, grasp it near one end, and try swinging it like a baseball bat.
Then grasp it in the center and rotate it as far as you can without hitting your body.
How does the rotational inertia with the axis at the end compare with the rotational inertia with the axis at the center? What is the ratio of the two?
Solution:
Although this may not be a good approximation, we will use the formulas for a thin rectangular plate of height h and width z from http://en.wikipedia.org/wiki/List_of_moments_of_inertia . (A 2x4 isn't very thin so this may not be a good approximation.)
When rotated around the end,
Iend = (1/3)*(m*h^2) + (1/12)*(m*z^2)
When rotated around the center,
Icen = (1/12)*(m*h^2 + m*z^2)
where
Define the numeric values
Let h = 96 inches and z = 3.75 inches (a 2x4 really isn't 2 inches thick and 4 inches wide)
Let mass = 1kg. We don't know what the mass of a 2.4 is. However, it will cancel out when we compute the ratio of the two cases. We can use any value so long as we don't ascribe any credibility to the absolute rotational inertia value.
Substitute numeric values for symbols
Iend = (1/3)*(1kg*(96inches)^2) + (1/12)*(1kg*(3.75 inches)^2)
Plugging this expression into the Google calculator gives us:
Iend = 1.98 m^2 kg
(Remember, however, that this isn't an accurate absolute value because we aren't using the actual mass of a piece of 2x4 lumber.)
Icen = (1/12)*(1kg*(96inches)^2 + 1kg*(3.75inches)^2)
The Google calculator gives us
Icen = 0.496 m^2 kg
And the ratio is...
If I formulated the problem correctly before plugging the expressions into the Google calculator, the ratio
Iend/Icen = (1.98 m^2 kg)/(0.496 m^2 kg) = 3.99
Therefore, it should have been about four times as difficult to swing the 2x4 like a baseball bat than to spin it at its center. (The downward torque caused by gravity probably made it seem even worse that that.)
Among other things, this scenario illustrates the parallel axis theorem .
Consider a dumbbell, or a barbell, whichever you choose to call it. This object consists of two identical solid spheres, each with mass M. The centers of mass of the spheres are separated by a distance L. The radius of each sphere is R.
The spheres are connected by a thin rod with mass m of length d. Thus, the length of the rod is L-2*R.
Find the rotational inertia of the dumbbell about an axis at the center of and perpendicular to the rod.
Solution:
The total rotational inertia of the dumbbell about the chosen axis consists of the sum of three parts:
There are really five items in the sum
We learned from the parallel axis theorem that the first two items in the above list are each made up of the sum of two items:
We learned from the earlier Examples of rotational inertia that the rotational inertia of a solid sphere about an axis through its center of mass is
Isphere = (2/5)*M*R^2
We learned in Rotational inertia that the rotational inertia of a point mass rotating about a chosen axis is
I = M*r^2, or in this case
I = M*(L/2)^2
Thus, the rotational inertia of each sphere about the chosen axis is the sum of those two, or
Isphere_axis = (2/5)*M*R^2 + M*(L/2)^2
The total rotational inertia of the dumbbell will be twice this value plus the rotational inertia of the rod about the chosen axis. We learned in Examples of rotational inertia that the rotational inertia about the chosen axis for the rod is
Irod = (1/12)*m*d^2
This, the total rotational inertia of the dumbbell about the chosen axis is
Itotal = Irod + 2*Isphere_axis, or
Itotal = (1/12)*m*d^2 + 2*((2/5)*M*R^2 + M*(L/2)^2)
Because the length of the rod, d is
d = L-2*R
We could substitute this expression for d giving us
Itotal = (1/12)*m*(L-2*R)^2 + 2*((2/5)*M*R^2 + M*(L/2)^2)
which reduces the expression down to include
I will leave it as an exercise for the student to assign typical numeric values to each of those variables and to compute the rotational inertia of a dumbbell.
Another good exercise for the student would be to replace the spheres with disks having the same mass and compare the rotational inertia of the two configurations.
In this scenario, a friction-free pulley with a mass M and a rotational inertia I is suspended from a beam. A cord is threaded around the pulley and two objects with masses of m1 and m2 are fastened to the ends of the cord.
The cord will not stretch and will not slip on the surface of the pulley.
When the two objects are held at the same level with the cord taut and released simultaneously, one may move up and the other may move down, depending of the relative mass values of the two objects. Because the cord cannot stretch, the magnitude of the velocity of each object must be the same.
Ignoring air resistance, write a general equation for the magnitude of the velocity of each object after each object has moved a distance h.
Solution:
With no friction and no air resistance, all forces acting on the system are conservative. Therefore, the mechanical energy of the system must be preserved.
deltaU + deltaK = 0
where
Since the pulley mechanism is in equilibrium, the only potential energy possessed by the system that can change is the gravitational potential energy of the two objects. Therefore, when the objects move, the gravitational potential energy of the two objects is converted into translational kinetic energy of the objects and rotational kinetic energy of the pulley.
Changes in potential energy of the system
In order to keep our algebraic signs straight, we will assume that m1 is greater than (or possibly equal to) m2. As a result, when the objects move, m1 will move down and m2 will move up.
This will result in the following changes in the potential energy of the system.
deltaU1 = -m1*g*h
deltaU2 = +m2*g*h
where
The mechanical energy of the system
The mechanical energy of the system includes the translational kinetic energy of each of the two objects plus the rotational kinetic energy of the pulley.
deltaK = (1/2)*(m1+m2)*v^2 + (1/2)*I*w^2
where
Translational speed versus tangential speed
Because the cord cannot slip on the pulley, the tangential speed of a point on the edge of the pulley must be equal to the translational speed of the objects.
This tangential speed at the edge of the pulley is related to the angular velocity of the pulley as follows
v = w*R, or
w = v/R
Where
Substitute, combine terms, and simplify
Substitution of v/R for w yields
deltaK = (1/2)*(m1+m2)*v^2 + (1/2)*I*(v/R)^2
Combining potential and kinetic energy yields
deltaU + deltaK = -m1*g*h + m2*g*h
+ (1/2)*(m1+m2)*v^2 + (1/2)*I*(v/R)^2 = 0
Simplification yields
(1/2)*(m1+m2)*v^2 + (1/2)*I*(1/R^2)*v^2 = (m1 - m2)*g*h
Solving for v^2 yields
v^2 = (2*(m1 - m2)*g*h)/((m1+m2) + I*(1/R^2)), or
The general equation for velocity is
v = ((2*(m1 - m2)*g*h)/((m1+m2) + I*(1/R^2)))^(1/2)
Make the following assumptions:
The pulley is a uniform disk with a mass, M, of 1 kg and a radius, R, of 1 meter.
m1 = 2 kg
m2 = 1 kg
h = 0.5 m
Find the velocity v.
Solution:
Referring back to Examples of rotational inertia , and changing the notation to match that being used in this scenario, we find that the rotational inertia for the pulley is
I = (1/2)*M*(R)^2
The rotational inertia
Let's begin by computing the rotational inertia of the pulley.
I = (1/2)*1kg*(1m)^2, or
I = 0.5*(m^2)*kg
Substitute values other than rotational inertia
Substituting values into the general equation yields
v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + I*(1/(1m)^2)))^(1/2)
where we still have the rotational inertia as a variable.
We will use this equation again later in this module.
Substitute the value for rotational inertia
Substituting the value for rotational inertia computed above gives us
v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + (0.5*(m^2)*kg)*(1/(1m)^2)))^(1/2)
We will also use this equation again later in this module.
Assuming that I managed to get all of that done without making an error, the velocity is
v = 1.67 m/s
Assume the same conditions as in part 2 , except that m2 = m1 = 2 kg, what is the velocity?
Solution:
Starting with the equation from above and replacing each occurrence of 1kg with 2kg yields
v = ((2*(2kg - 2kg)*(9.8m/s^2)*0.5m)/((2kg+2kg) + (0.5*(m^2)*kg)*(1/(1m)^2)))^(1/2)
Using the Google calculator to solve this equation tells us that
v = 0 m/s.
This is what we should expect for the two objects having the same mass. Each object has an equal desire to fall toward the earth, so they balance one another, causing the system to be in equilibrium.
Go back to the conditions for part 2 except instead of assuming that the pulley is a uniform disk, assume that the pulley approximates a "thick-walled cylindrical tube with open ends, of inner radius r1, and outer radius r2" .
For this configuration, we can approximate the rotational inertia as
I = (1/2)*m*(r1^2 + r2^2) (see Examples of rotational inertia )
For example, think of a pulley that looks something like a bicycle wheel with very lightweight spokes connecting the outer rim to the axle.
Let r1 = 0.9*r2 and let the mass be unchanged.
Find the velocity.
Solution:
Given that the inner radius is 0.9 times the outer radius and the mass is unchanged, we can approximate the rotational inertia for this pulley with
I = (1/2)*M*((0.9*R)^2 + R^2)
The rotational inertia
Let's compute the rotational inertia for this pulley.
I = (1/2)*1kg*((0.9*1m)^2 + (1m)^2), or
I = 0.905 (m^2)*kg
As you can see, the rotational inertia for this configuration is almost double the rotational inertia for the uniform disk configuration.
Start with the original equation
Go back and get the original equation that contains numeric values but still has the rotational inertia as a variable.
v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + I*(1/(1m)^2)))^(1/2)
Substitute the rotational inertia
Replace the rotational inertia, I, with the approximate rotational inertia for our new pulley.
v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + (0.905 (m^2)*kg)*(1/(1m)^2)))^(1/2)
Once again, if I managed to get through all of that without making an error, the velocity is
v = 1.58 m/s
Analysis
Comparing this value with the velocity from part 2 , , we see that the velocity was reduced from 1.67 m/s to 1.58 m/s due to the increase in the rotational inertia of the pulley.
This is what we should expect. Increasing the rotational inertia of the pulley causes it to take longer to accelerate given the same external force that is causing it to change its angular velocity and acquire rotational kinetic energy.
Let's make one more adjustment to the geometry of the pulley and observe the effect that it has on the system.
Keep all of the parameters the same as part 4 except let the radius of the pulley, R, be 10m.
Find the velocity.
Solution:
Begin by computing the rotational inertia of the new pulley.
I = (1/2)*M*((0.9*R)^2 + R^2), or
I = (1/2)*1kg*((0.9*10m)^2 + (10m)^2), or
I = 90.5 (m^2)*kg
Note that the rotational inertia is proportional to the square of the radius. Therefore, increasing the radius by a factor of 10 caused the rotational inertia to be increased by a factor of 100.
Generally speaking, moving the concentration of mass further from the axis of rotation will increase the rotational inertia.
Start with the original equation as before
Going back and getting the original equation that contains numeric values but still has the rotational inertia as a variable, we have.
v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + I*(1/(1m)^2)))^(1/2)
Substitute the new rotational inertia
Replacing the rotational inertia, I, with the rotational inertia for our new pulley yields.
v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + (90.5 (m^2)*kg)*(1/(1m)^2)))^(1/2)
Solving this equation with the Google calculator tells us that the new velocity value is
v = 0.323 m/s
which is much lower than the velocity for either part 2 or part 4 .
In case you want to learn more about this topic, the arrangement of the pulley and the objects that we have been discussing is commonly called an Atwood machine. The device was invented in 1784 by Rev. George Atwood as a laboratory experiment to verify the mechanical laws of motion.
A flywheel is a device that is commonly used to smooth out the irregularities of angular velocity in rotational motion. For example, in the age of steam engines, the rotational motion was powered only during a portion of each cycle, and was coasting during the remainder of the cycle. Unless corrected, therefore, a loaded steam engine output shaft would speed up and slow down once during each rotation of the shaft.
The correction
Most steam engines employed a large flywheel that was turned by the output shaft to cause the angular velocity to remain relatively constant despite the fact that rotational power was applied during only a portion of the cycle.
When rotational power was applied to the output shaft, the shaft turned the flywheel. When rotational power was not applied to the output shaft, the flywheel turned the shaft. Because the flywheel had a large rotational inertia, it preferred to turn at a near constant angular velocity.
Physically, the flywheels were usually large wheels with spokes and most of the mass distributed in a rim at the circumference of the wheel. This configuration produced a large rotational inertia for a given amount of mass and a given amount of available space.
Perhaps the most important thing for you to take away from this module is that
Two rotating objects having exactly the same mass can have entirely different rotational inertia values (moments of inertia).
For example, a flywheel with the bulk of its mass concentrated a large distance from the axis of rotation is much more effective in smoothing out the angular velocity of the device than would be a flywheel with the same mass concentrated in a small radius near the axis of rotation.
I encourage you to work through the examples that I have presented in this lesson to confirm that you get the same results. Experiment with the examples, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.
I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.
This section contains a variety of miscellaneous information.
Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.
I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.
Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.
-end-
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Last edited by Richard Baldwin on Oct 8, 2012 1:39 pm -0500.
"Blind students should not be excluded from physics courses because of inaccessible textbooks. The modules in this collection present physics concepts in a format that blind students can read […]"